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Question
The cell in which the following reactions occurs:
\[\ce{2Fe^{3+}_{( aq)} + 2I^-_{ (aq)} -> 2Fe^{2+}_{( aq)} + I2_{(s)}}\] has \[\ce{E^\circ_{cell}}\] = 0.236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
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Solution
\[\ce{2Fe^{3+} + 2e- -> 2Fe^{2+}}\]
\[\ce{2I- -> I2 + 2e-}\]
________________________________________
\[\ce{2Fe^{3+} + 2I^- -> 2Fe^{2+} + I2}\]; \[\ce{E^\circ_{cell}}\] = 0.236 V
ΔG° = −nFE°
= −2 × 0.236 × 96500
= −45548 J mol−1
= −45.55 kJ mol−1
ΔG° = −2.303 RT log K
log K = \[\ce{-\frac{\Delta G^\circ}{2.303 RT}}\]
= `(-(-45548))/(2.303 xx 8.314 xx 298)`
= `(-(-45548))/5705.84`
log K = 7.98
K = antilog 7.98
= 9.54 × 107
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